algebra

Álgebra Linear – Prof. Mario Balistieri Sobrinho

Engenharia de Controle e Automação
Paulo Cesar de Aquino RA 4649890440 Turma A

Resolução da lista de exercícios

Resolução
01) a=A.B
A= B= a11=1.1+3.(-2)+2.0=-5 a21=2.1+1.(-2)+1.0=0 a31=(-3).1+(-1).(-2)+0.3=-5 a12=1.2+3.3+2.4=19 a22=2.2+1.3+1.4=11 a32=(-3).2+(-1).3+0.4= -11 a13=1.3+3.(-4)+2.2=-5 a23=2.3+1.(-4)+1.2=4 a33=(-3).3+(-1).(-4)+0.2= -5

_____________________________________________________________________________ b= B.A
B= A= b11=1.1+2.2+3.(-3)=-4 b21= (-2) .1+3.2+(-4).(-3)=16 b31=0.1+4.2+2.(-3)=2 b12=1.3+2.1+3.(-1)=2 b22=(-2).3+3.1+(-4).(-1)=1 b32=0.3+4.1+2.(-1)=2 b13=1.2+2.1+3.0= 4 b23=(-2).2+3.1+(-4).0=-1 b33=0.2+4.1+2.0= 4 _____________________________________________________ c=A.(C.B) A= B= C=
C.B=
C.B 11=2.1+3.(-2)+1.0=-4 C.B 21= 5.1+0.(-2)+2.0=5 C.B 31=1.1+1.(-2)+0.0=-1
C.B 12=2.2+3.3+1.4=17 C.B 22=5.2+0.3+2.4=18 C.B 32=1.2+1.3+0.4=5
C.B 13=2.3+3.(-4)1.2= -4 C.B 23=5.3+0.(-4)2.2=19 C.B 33=1.3+1(-4)+0.2=-1
A= C.B=
A.(C.B)= n n11=1.(-4)+3.5+2(-1)=9 n21=2.(-4)+1.5+1.(-1)=-4 n31=(-3).(-4)+(-1).5+0.(-1)=7 n12=1.17+3.18+2.5=81 n22=2.17+1.18+1.5=57 n32=(-3).17+(-1).18+0.5=-69 n13=1.(-4)+3.19+2.(-1)=51 n23=2.(-4)+1.19+1.(-1)=10 n33=(-3).(-4)+(-1).19+0.1=-7 _____________________________________________________________________ d) X = 3.A + 2.B – 2.C
A= B= C=
X=3. + 2. -2. x= + - x= - x= _____________________________________________________________________ 2 ) Considerando as matrizes do exercício anterior, calcular:
a) Det. A b) Det. A.B c) Det. C
a)

Det. A== =